From: Albert_Rich@msn.com   
      
   On Wednesday, April 26, 2017 at 9:09:43 AM UTC-10, clicl...@freenet.de wrote:   
   > "Nasser M. Abbasi" schrieb:   
   > >   
   > > On 4/26/2017 5:15 AM, Albert Rich wrote:   
   > > >   
   > > > An antiderivative of arctanh(t)*log(t)/(t-1) wrt t is:   
   > > >   
   > > > -1/2*Log[2]*Log[t]*Log[(1 + t)/2] + 1/4*Log[2]*Log[(1 + t)/2]^2 +   
   > > >  1/2*Log[1 - t]*Log[t]*Log[1 + t] -   
   > > >  1/2*Log[2]*Log[(1 + t)/2]*Log[(1 + t)/(2*t)] +   
   > > >  1/4*Log[1 - t]*Log[(1 + t)/(2*t)]^2 +   
   > > >  1/4*Log[-(1/t)]*Log[(1 + t)/(2*t)]^2 -   
   > > >  1/4*Log[(-1 + t)/(2*t)]*Log[(1 + t)/(2*t)]^2 +   
   > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, 1 + 1/t] -   
   > > >  ArcTanh[t]*PolyLog[2, 1 - t] +   
   > > >  1/2*Log[1 + t]*PolyLog[2, 1 - t] + 1/2*Log[1 + t]*PolyLog[2, t] -   
   > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, t] +   
   > > >  1/2*Log[t]*PolyLog[2, (1 + t)/2] +   
   > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, (1 + t)/2] -   
   > > >  1/2*Log[(1 + t)/(2*t)]*PolyLog[2, (1 + t)/(2*t)] -   
   > > >  1/2*PolyLog[3, 1 + 1/t] -   
   > > >  1/2*PolyLog[3, 1 - t] - (1/2)*PolyLog[3, t] -   
   > > >  1/2*PolyLog[3, (1 + t)/2] +   
   > > >  1/2*PolyLog[3, (1 + t)/(2*t)]   
   > > >   
   > > > Taking the difference of evaluating its limits as t approaches 1   
   > > > and 0 gives 1/8 pi^2 log(2).   
   > >   
   > > Amazing. Well done. I assume you found a rule for this and it   
   > > will be in next Rubi version.   
   >   
   > Perhaps a better nose when and how to integrate by parts is all that's   
   > needed?   
      
   Yes, for integrands involving inverse elementary functions, integration by   
   parts is obviously the way to go. The following partial derivation of Nasser's   
   integral uses parts twice:   
      
   Int[ArcTanh[x]*Log[x]/(x-1),x]   
      
   == -ArcTanh[x]*PolyLog[2,1-x] +   
      Int[PolyLog[2,1-x]/(1-x^2),x]   
      
   == -ArcTanh[x]*PolyLog[2,1-x] +   
      1/2*Int[PolyLog[2,1-x]/(1-x),x] +   
      1/2*Int[PolyLog[2,1-x]/(1+x),x]   
      
   == -ArcTanh[x]*PolyLog[2,1-x] -   
      1/2*PolyLog[3,1-x] +   
      1/2*Int[PolyLog[2,1-x]/(1+x),x]   
      
   == -ArcTanh[x]*PolyLog[2,1-x] -   
      1/2*PolyLog[3,1-x] +   
      1/2*Log[1+x]*PolyLog[2,1-x] +   
      1/2*Int[Log[x]*Log[1+x]/(x-1),x]   
      
   Then I got stuck.  Fortunately however, Mathematica's built-in integrator   
   could integrate log(x)*log(1+x)/(x-1), although for some reason it could not   
   integrate the original problem.   
      
   So if somebody has a better nose than I how to step through the integration of   
   log(a+b*x)*log(c+d*x)/(e+f*x), Rubi and I would very much like to know...   
      
   Albert   
      
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